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4 years ago

Can someone plz help me solved this problem! I’m giving you 10 points! I need help plz help me! Will mark you as brainiest!

Mathematics

1 answer:

pickupchik [31]4 years ago

7 0

Hey there! :)

Answer:

a. 3

b. -22

c. -2

d. -2

e. 5a + 8

f. a² + 6a + 3

Step-by-step explanation:

Calculate the answers by substituting the values inside of the parenthesis for 'x':

a. f(1) = 5(1) - 2 = 3

b. f(-4) = 5(-4) - 2 = -22

c. g(-3) = (-3)² + 2(-3) - 5 = 9 - 6 - 5 = -2

d. g(1) = 1² + 2(1) - 5 = 1 + 2 -5 = -2

e. f(a+ 2) = 5(a+2) - 2 = 5a + 10 - 2 = 5a + 8

f. g(a + 2) = (a + 2)² + 2(a + 2) - 5 = a² + 4a + 4 + 2a + 4 - 5 =

a² + 6a + 3

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Solve the following differential equation using using characteristic equation using Laplace Transform i. ii y" +y sin 2t, y(0) 2

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Answer:

The solution of the differential equation is $y(t)= - \frac{1}{3} Sin(2t)+2 Cos(t)+\frac{5}{3} Sin(t)$

Step-by-step explanation:

The differential equation is given by: y" + y = Sin(2t)

<u>i) Using characteristic equation:</u>

The characteristic equation method assumes that y(t)= $e^{rt}$ , where "r" is a constant.

We find the solution of the homogeneus differential equation:

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As $e^{rt}$ could never be zero, the term (r²+1) must be zero:

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$y(t)_{h}=c_{1}e^{it}+c_{2}e^{-it}$

Using Euler's formula:

$y(t)_{h}=c_{1}[Sin(t)+iCos(t)]+c_{2}[Sin(t)-iCos(t)]$

$y(t)_{h}=(c_{1}+c_{2})Sin(t)+(c_{1}-c_{2})iCos(t)$

$y(t)_{h}=C_{1}Sin(t)+C_{2}Cos(t)$

The particular solution of the differential equation is given by:

$y(t)_{p}=ASin(2t)+BCos(2t)$

$y'(t)_{p}=2ACos(2t)-2BSin(2t)$

$y''(t)_{p}=-4ASin(2t)-4BCos(2t)$

So we use these derivatives in the differential equation:

$-4ASin(2t)-4BCos(2t)+ASin(2t)+BCos(2t)=Sin(2t)$

$-3ASin(2t)-3BCos(2t)=Sin(2t)$

As there is not a term for Cos(2t), B is equal to 0.

So the value A=-1/3

The solution is the sum of the particular function and the homogeneous function:

$y(t)= - \frac{1}{3} Sin(2t) + C_{1} Sin(t) + C_{2} Cos(t)$

Using the initial conditions we can check that C1=5/3 and C2=2

<u>ii) Using Laplace Transform:</u>

To solve the differential equation we use the Laplace transformation in both members:

ℒ[y" + y]=ℒ[Sin(2t)]

ℒ[y"]+ℒ[y]=ℒ[Sin(2t)]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]-s·y(0)-y'(0)=s²·Y(s) -2s-1

ℒ[y]=Y(s)

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We replace the previous data in the equation:

s²·Y(s) -2s-1+Y(s) = $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)-2s-1= $\frac{2}{(s^{2}+4)}$

(s²+1)·Y(s)= $\frac{2}{(s^{2}+4)}+2s+1=\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)}$

Y(s)= $\frac{2+2s(s^{2}+4)+s^{2}+4}{(s^{2}+4)(s^{2}+1)}$

Y(s)= $\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}$

Using partial franction method:

$\frac{2s^{3}+s^{2}+8s+6}{(s^{2}+4)(s^{2}+1)}=\frac{As+B}{s^{2}+4} +\frac{Cs+D}{s^{2}+1}$

2s^{3}+s^{2}+8s+6=(As+B)(s²+1)+(Cs+D)(s²+4)

2s^{3}+s^{2}+8s+6=s³(A+C)+s²(B+D)+s(A+4C)+(B+4D)

We solve the equation system:

A+C=2

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